Answer:
The total time that Jim needs to change x oil changes and y tire changes is less than 180 min.
The time needed for x oil changes is 12 * x.
The time needed for y tire changes is 18 * y.
The total time is the sum of the above times and needs to be less than 180 that is
12 * x + 18 * y < 180 divide both sides of equation by 6
12/6 * x + 18/6*y < 180/6
2*x + 3*y < 30
2*x < 30 - 3*y divide both sides by 2 to get the inequality for x
x < 30/2 - 3/2*y = 15 - 1.5 y < 15 that is x < = 15
2*x + 3*y < 30
3*y < 30 - 2*x divide both sides by 3 to get the inequality for y
y < 30/3 - 2/3 *x = 10 - 2/3*x < 10 that is y < = 10
Also we can write x + y < x+ 3/2 * y < 15.
Explanation:
Jim's can do not more then 5 oil changes and not more then 10 tire changes or all together she can do not more then 15 total of oil and tire changes.
Answer:
This question appears incomplete
Explanation:
However, it should be noted that addition of soluble salts generally lowers the freezing point of water hence after the addition, water will no longer freeze at 0°C but lower.
Soluble salts tend to form more ions in water, it is these ions that are responsible for interfering with the hydrogen bonds hence lowering the freezing. Thus, (since each bag are of the same weight) <u>the bag that contains the salt that ionizes more in water will lower the freezing point by the greatest amount</u>.
NOTE: Different weight of the salts could lead to more ions been formed in the water by some salts as against the other.
We need to measure 20.0 grams of CaCl₂ to prepare 500 mL of 0.360 M solution.
First, we need to determine the required moles of CaCl₂. We have 500 mL (0.500 L) of a 0.360 M solution (0.360 moles of CaCl₂ per liter of solution).
Then, we will convert 0.180 moles to grams using the molar mass of CaCl₂ (110.98 g/mol).
To prepare the solution, we weigh 20.0 g of CaCl₂ and add it to a beaker with enough distilled water to dissolve it. We stir it, heat it if necessary, and when we have a solution, we transfer it to a 500 mL flask and complete it to the mark with distilled water.
We need to measure 20.0 grams of CaCl₂ to prepare 500 mL of 0.360 M solution.
You can learn more about solutions here: brainly.com/question/2412491
